The correct option is D 78 Lets reduce the fraction as follows: 2^n+4 2^n+1 #160; #160;2^n+4 lets take 2^n+1 common in numerator and denom ...

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nCr Definitions and Properties

2n+4-22n22n+3...

Question

$\frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})} =$ ____.

2^{n + 1} - \frac{1}{8}

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- 2^{n + 1}

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1 - 2^{n}

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\frac{7}{8}

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Solution

The correct option is D $\frac{7}{8}$
Lets reduce the fraction as follows:
$\frac{2^{n + 4} - 2^{n + 1}}{2^{n + 4}}$

lets take $2^{n + 1}$ common in numerator and denominator,

$\frac{2^{n + 1} (2^{3} - 1)}{2^{n + 1} (2^{3})}$

$\frac{2^{3} - 1}{2^{3}} = 7 / 8$

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(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},

\sum \sum 0 \leq r < s \leq n (r + s) C_{r} C_{s}

\frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})}

\frac{1^{2}}{(1) (3)} + \frac{2^{2}}{(3) (5)} + . . . . . + \frac{n^{2}}{(2 n - 1) (2 n + 1)} = \frac{n (n + 1)}{2 (2 n + 1)}

\frac{1}{(1) (3)} + \frac{1}{(3) (5)} + . . . . . + \frac{1}{(2 n - 1) (2 n + 1)} = \frac{1}{2 n + 1}

\frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})} + 2^{- 3}

\frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})} + 2^{- 3}

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